Interactive Applet |
You can move the points A, B and C (click on the point and drag it).
Press the keys “+” and “−” to zoom in or zoom out the visualization window and use the arrow keys to translate it.
You can also construct all centers related with this one (as described in ETC) using the “Run Macro Tool”. To do this, click on the icon _{}, select the center name from the list and, then, click on the vertices A, B and C successively.
Information from Kimberling's Encyclopedia of Triangle Centers |
Trilinears csc A tan A/4 : csc B tan B/4 : csc C tan C/4
Barycentrics tan A/4 : tan B/4 : tan C/4
= f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = (b + c - a)^{1/2}[2b^{1/2}c^{1/2} - (a + b + c)^{1/2}(b + c - a)^{1/2}]Along each side of ABC there is a segment that is a common tangent to two of the three Malfatti circles of ABC. Let A', B', C' be the midpoints of these respective segments. Then triangle A'B'C' is perspective to ABC, and the perspector is X(1143). (Stanley Rabinowitz, #4611, 12/29/01) For coordinates, see Paul Yiu, #4615, 12/30/01, and
Milorad R. Stevanovic, "Triangle Centers Associated with the Malfatti Circles," Forum Geometricorum 3 (2003) 83-93.
X(1143) lies on these lines: 8,177 174,175 558,1488
X(1143) = isotomic conjugate of X(1274)